In question number 91, maximum value of magnetisation of the given domain is (Dipole moment of an iron atom `9.27 xx 10^(-24)A m^(2))`

Options

A. `8.0xx10^(5)Am^(-1)`
B. `6.0xx10^(4)Am^(-1)`
C. `8.0xx10^(3)Am^(-1)`
D. `6.0xx10^(3)Am^(-1)`

Explanation

To solve the problem of finding the maximum value of magnetization of a given domain, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- The dipole moment of an iron atom \( m = 9.27 \times 10^{-24} \, \text{A m}^2 \).

2. **Determine the Number of Atoms**:
- We need to assume or find the number of atoms (or dipoles) in the domain. For this problem, we will use the standard value \( N = 6.9 \times 10^{11} \) atoms.

3. **Calculate the Net Dipole Moment**:
- The net dipole moment \( M_n \) can be calculated using the formula:
\[ M_n = N \times m \]
- Substituting the values:
\[ M_n = 6.9 \times 10^{11} \times 9.27 \times 10^{-24} \]
- Performing the multiplication:
\[ M_n = 6.4 \times 10^{-12} \, \text{A m}^2 \]

4. **Calculate the Net Magnetization**:
- The net magnetization \( M \) is given by the formula:
\[ M = \frac{M_n}{V} \]
- For a standard volume \( V = 8 \times 10^{-18} \, \text{m}^3 \):
\[ M = \frac{6.4 \times 10^{-12}}{8 \times 10^{-18}} \]
- Performing the division:
\[ M = 8 \times 10^{5} \, \text{A/m} \]

5. **Final Result**:
- The maximum value of magnetization of the given domain is:
\[ M = 8 \times 10^{5} \, \text{A/m} \]